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• 1. Unsteady State Conduction Evelyn R. Laurito Lani Pestano Ch.E. 206
• 2. Lecture Objectives
• To understand the concept of unsteady state conduction
• To study the case of unidirectional unsteady state conduction
• To understand how to use Geankoplis Charts in solving unidirectional unsteady state conduction problems
• Gurney and Lurie Charts
• Heisler Chart
• Chart for Average Temperature
• Chart for Semiinfinite solid
• To understand how to use Numerical Methods in solving unidirectional unsteady state conduction problems
• This happens when the temperature gradient across the solid changes with time.
• This may be due to unstable boundary temperatures at startup, sudden temperature fluctuations during steady state conditions, or internal generation of heat.
• Sample Cases:
• Startup of a Furnace
• Heat Treatment of Solids
• Deep Oil Frying
• Change of Weather
• 4. Unidirectional Unsteady State Case  x  y  z x x+  x q in q out A = V = Solid properties:  , c P Mass = Heat Balance Across  x: q in - q out = Rate of heat accumulation Rate of heat accumulation = Using Fourier’s Law: q in = q out = c P  y  z  x  y  z  x  y  z   x  y  z   T  t - k  y  z  T  x x - k  y  z  T  x x +  x
• 5. Unidirectional Unsteady State Case The Heat Balance becomes: - = Simplifying: -  x From Calculus: Final Equation:  but - k  y  z  T  x x - k  y  z  T  x x +  x  x  y  z  c P  T  t  T  t = k  c P  T  x x +  x  T  x x  T  x x +  x -  T  x x  x =  2 T  x 2  T  t = k  c P =  2 T  x 2 
• 6. Unidirectional Unsteady State Case
• Depends on solid geometry
• Requires PDE solution methods that results into Fourier series solutions that are tedious to evaluate
• May be simplified by the use of charts or numerical methods
Use of Charts:
• Gurnie-Lurie Charts – to determine point temperatures
• Heisler Charts – to determine central temperatures
• Average Temperature Chart
• Chart for Semiinfinite Solids
 T  t =  2 T  x 2  Solution of:
• 7. Geankoplis Charts
• Gurney-Lurie Charts
• Fig. 5.3-5/340 for large flat plate
• Fig. 5.3-7/343 for long cylinder
• Fig. 5.3-9/345 for sphere
• Heisler Chart
• Fig. 5.3-6/341 for large flat plate
• Fig. 5.3-8/344 for long cylinder
• Fig. 5.3-10/346 for sphere
• Fig. 5.3-13/349 for Ave. Solid Temperature
• Fig. 5.3-3/337 for Semi-infinite solid
• 8. Nomenclature
• Gurney-Lurie and Heisler Charts:
• T o = temperature at t(time)= 0 (uniform)
• T 1 = new and constant surface temperature
• x 1 = ½ plate thickness, outer radius of cylinder or sphere
•  = constant thermal diffusivity
• X =  t/ x 1 2 : relative time
• x = distance from plate center or any radius of a cylinder or a sphere
• n = x/x 1 : relative position
• T = point temperature at position x and time t
• Y = (T 1 -T)/(T 1 -T o ) :unaccomplished temp. change
• h = convective heat transfer coefficient
• m = k/(hx 1 ) : relative resistance
• 9. Nomenclature
• Average Temperature Chart
• T o = temperature at t(time)= 0 (uniform)
• T 1 = new and constant surface temperature
• T av = average solid temperature at time t
• E = (T 1 -T av )/(T 1 -T o )
• a = ½ plate thickness, outer radius of cylinder or sphere
• b = ½ plate width
• c = ½ plate length, ½ cylinder length
• 10. Nomenclature
• Chart for Semi-infinite solid
• Semi-infinite solid – solid where the unidirectional conductive heat transfer is infinite (Ex. Ground)
• T o = initial uniform solid temperature
• T 1 = constant ambient temperature to which solid surface is exposed
• T = temperature of solid at position x
• 1- Y = (T-T o )/(T 1 -T o ): Ordinate
• h(  t) 0.5 /k : convective parameter
• x/[2 (  t) 0.5 ]: Abscissa
• 11. Problems from Geankoplis
• Exercises:
• 5.3-5 (Plate)
• 5.3-7 (Long Cylinder)
• 5.3-9 (Sphere)
• Find the average solid temperature for all of the above cases
• 5.3-3
• Homework:
• Geankoplis: 5.3-2;5.3-4;5.3-6;5.3-8;5.3-10 (pages 375-377)
• Foust: 11.18;11.20;11.21(page 231)
• 12. 5.3-5 Cooling a Slab of Meat
• A slab of meat 25.4 mm thick originally at a uniform temperature of 10 o C is to be cooked from both sides until the center reaches 121 o C in an oven at 177 o C. The convection coefficient can be assumed constant at 25.6 W/m 2 -K. Neglect any latent heat changes and calculate the time required. The thermal conductivity is 0.69 W/m-K and the thermal diffusivity 5.85x10 -4 m 2 /h.
• 13. Solution for 5.3-5
• Given:
• t= 25.4 mm
• T o =10 o C
• T=121 o C (Center T) at x 1
• T 1 =177 o C
• h=25.6 W/m 2 -K
• k=0.69 W/m-K
•  =5.85x10 -4 m 2 /h
• Required: t= ?
• 14. Solution for 5.3-5
• x 1 =12.7mm
• x=25.4mm
• 15. 5.3-7 Cooling of a Steel Rod
• A long steel rod 0.305 m in diameter is initially at a temperature of 588K. It is immersed in an oil bath maintained at 311K. The surface convective coefficient is 125 W/m 2 -K. Calculate the temperature at the center of the rod after 1 h. The average physical properties of the steel are k=38 W/m-K and  =0.0381m 2 /h
• 16. Solution for 5.3-7
• Given:
• D= 0.305 m
• T o =588 K
• T 1 =311 K
• h=125 W/m 2 -K
• t=1 h
• k=38 W/m-K
•  =0.0381 m 2 /h
• Required: T at the center
• 17. Solution for 5.3-7
• 18. 5.3-9 Temp. of Oranges on Trees During Freezing Weather
• In orange-growing areas, the freezing of the oranges on the trees during cold nights is economically important. If the oranges are initially at a temperature of 21.1 o C, calculate the center temperature of the orange if exposed to air at –3.9 o C for 6 h. The oranges are 102 mm in diameter and the convective coefficient is estimated as 11.4W/m 2 -K. The thermal conductivity k is 0.431 W/m-K and  =4.65x10 -4 m 2 /h. Neglect any latent heat effects.
• 19. Solution for 5.3-9
• Given:
• D= 102 m  x=102/2=51mm
• T o =21.1 o C=294.1K
• T 1 =-3.9 o C=269.1K
• h=11.4 W/m 2 -K
• t=6 h
• k=0.431 W/m-K
•  =4.65x10 -4 m 2 /h
• Required: T at the center
• 20. Solution for 5.3-9 From Fig. 5.3-10: Y = 0.05
• 21. 5.3-3 Cooling a Slab of Aluminum
• A large piece of aluminum that can be considered a semi-infinite solid initially has a uniform temperature of 505.4K. The surface is suddenly exposed to an environment at 338.8K with a surface convection coefficient of 455W/m 2 -K. Calculate the time in hours for the temperature to reach 388.8 K at a depth of 25.4 mm. The average physical properties are  =0.340m 2 /h and k=208W/m-K.
• 22. Solution for 5.3-3
• Given:
• D= 0.305 m
• T o =505.4 K
• T 1 =338.8 K
• T=388.8K when x=25.4mm
• h=455 W/m 2 -K
• k=208 W/m-K
•  =0.304 m 2 /h
• Required: time in hours for the temperature to reach 388.8K at a depth of 25.4 mm
• 23. Solution for 5.3-9 From Fig. 5.3-10: Y = 0.05
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